Contrapositives and Monotonic Functions

Date: 03/17/2003 at 12:09:24
From: B.Y. Vinay Kumar
Subject: Monotonic functions
We usually define an increasing function as (say)
if x1<=x2 implies f(x1) <=f(x2) then f is said to be increasing
BUT will it make any difference if we define it as
if f(x1) <=f(x2) implies x1<=x2 then f is increasing
I just wanted to know whether the converse holds.

Date: 03/18/2003 at 04:08:59
From: Doctor Jacques
Subject: Re: Monotonic functions
Hi,
There may be some confusion regarding the definition of an increasing
function, i.e. whether or not we consider a constant function as
increasing. To avoid ambiguity, we will only use "non-decreasing"
and "strictly increasing."
Your first definition describes a non-decreasing function, i.e. it is
also satisfied by a constant function. But the second definition is
not satisfied by a constant function: we always have f(x1) <= f(x2),
whatever x1 and x2 may be. You can, however, define a non-decreasing
function by:
f(x1) < f(x2) --> x1 < x2
In the case of a constant function, this is verified, since the LHS is
always false, and p -> q is true whenver p is false.
To describe a strictly increasing function, we could write:
x1 < x2 --> f(x1) < f(x2)
and this is equivalent to:
f(x1) <= f(x2) --> x1 <= x2
The logic behind these equivalences is based on the concept of
contraposition.
The contrapositive of a logic statement p --> q is (~q) --> (~p),
where ~ stands for negation. A statement and its contrapositive are
equivalent. You can see an example of this kind of reasoning in:
Contrapositive
http://mathforum.org/library/drmath/view/55706.html
Let us start with the first definition (non-decreasing function) :
x1 <= x2 --> f(x1) <= f(x2)
The contrapositive is:
~(f(x1) <= f(x2)) --> ~(x1 <= x2)
Now, in a totally ordered set, ~(a <= b) is the same as b > a. Our
statement becomes:
f(x1) > f(x2) --> x1 > x2
or
f(x2) < f(x1) --> x2 < x1
In this statement, x1 and x2 stand for any numbers, so we may
interchange the names of these variables. We get:
f(x1) < f(x2) --> x1 < x2
as shown before. The logic for a strictly increasing function is
entirely similar.
Note that, as x1 = x2 trivially implies f(x1) = f(x2), we could even
define a non-decreasing function by the (apparently) weaker statement:
x1 < x2 --> f(x1) <= f(x2)
which is equivalent to:
f(x1) < f(x2) --> x1 <= x2
Does this help? Write back if you'd like to talk about this some
more, or if you have any other questions.
- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/